Determine the order and commutativity of a group

Question

All matrices of the form

$\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$

where $a,b,c\in\mathbb{Z}/3\mathbb{Z}$, produce a group with respect to the matrix multiplication.

I want to show that every element (except the neutral one) is of order 3.

What I first did, was to compute

$\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}^3 = \begin{pmatrix} 1 & 3a & 3b +3ac \\ 0 & 1 & 3c \\ 0 & 0 & 1 \end{pmatrix}$

$\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 2a & 2b +2ac \\ 0 & 1 & 2c \\ 0 & 0 & 1 \end{pmatrix}$

Now it is obvious, since $3a \cong0\ \pmod 3$ and analogously for the other entries.

I assume this to be correct so far.

However, I wasn't able to show that this group is not commutative with respect to the matrix multiplication.

Answer 1

Try $$\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\begin{pmatrix}1&a'&b'\\0&1&c'\\0&0&1\end{pmatrix}$$

The upper right entry of the result is $b'+ac'+b$, so you need two matrices such that $ac'\neq a'c$.

Answer 2

Take $A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$ and $B=\begin{pmatrix}1&1&1\\0&1&0\\0&0&1\end{pmatrix}$.

Calculate $AB$ and $BA$. You will find $AB\neq BA$. Therefore, the group cannot be commutative.